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gladu [14]
3 years ago
8

A rifle with a mass of m1 = 3.4 kg fires a bullet with a mass of 8.3 g (0.0083 kg). The bullet moves with a velocity of 860 m/s

after the rifle is fired. What is the momentum of the bullet after the rifle is fired?
Physics
1 answer:
pogonyaev3 years ago
4 0

Answer:

P = 7.138 kg m/s

Explanation:

As we know that momentum is the product of mass and velocity

So here we have

m = 0.0083 kg

velocity = 860 m/s

so momentum of the bullet is given as

P = mv

P = (0.0083)(860)

P = 7.138 kg m/s

so the momentum of bullet after it has been shoot from the rifle is given as

P = 7.138 kg m/s

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What is wrong with the following statement: When you exert a force on a baseball, the equal and opposite force on the ball balan
gregori [183]

Answer:

When you exert a force on a baseball, there exists an equal and opposite force on the ball therefore, the ball will accelerate in opposite direction.

Explanation:

When you hit a ball with baseball bat, the bat exerts a great force on the ball which causes the ball to accelerate in the opposite direction. It is to be noted that the mass of bat is much greater than mass of ball but the acceleration of ball is also greater than the acceleration of the bat so both bat and ball almost exert same magnitude of force but in opposite direction and as a result both bat and ball accelerate in opposite direction, the deciding factor is of course the relative force applied by the batter and the bowler.

6 0
3 years ago
If a baseball has a zero velocity at some instant, is the acceleration of the baseball necessarily zero at that time? Explain -
ipn [44]

Answer:

No, not necessarily

Explanation:

If an object is moving with an acceleration that causes its speed to be reduced, there will be a moment in which it reaches v = 0, but this doesn't necessarily mean that the acceleration isn't acting anymore. If the object continues its movement with the same acceleration, it's velocity will become negative.

An example of an object that has zero velocity but non-zero acceleration:

If you throw an object in the air with a certain velocity, it will move vertically, reducing its velocity in a 9,8 m/s^{2} rate (which is the acceleration caused by gravity). At a certain point, the object will reach its maximum height, and will start to fall. In the exact moment that it reaches the maximum height, before it starts falling, its velocity is zero, but gravity is still acting on the object (this is the reason why it starts falling instead of just being stopped at that point). Therefore, at that point, the object has zero velocity but an acceleration of 9,8 m/s^{2}.

3 0
3 years ago
If an object accelerates from rest, with a constant acceleration of 5.4 m/s2, what will its velocity be after 28s?
aleksklad [387]
Vf = Vi + at
Vf = 0 + 5.4•28
= 151.2m/s..
not sure if its right
6 0
3 years ago
I am struggling on this physics question. Brainly is my last hope. Could somebody please provide an answer to this question, wit
Aleks [24]

1) 29.4 N

The force of gravity between two objects is given by:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M and m are the masses of the two objects

r is the separation between the centres of mass of the two objects

In this problem, we have

M=5.97\cdot 10^{24} kg (mass of the Earth)

m=3.0 kg (mass of the box)

r=R=6.37\cdot 10^6 m (Earth's radius, which is also the distance between the centres of mass of the two objects, since the box is located at Earth's surface)

Substituting into the equation, we find F:

F=\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24})(3.0)}{(6.37\cdot 10^6)^2}=29.4 N

2) g=9.8 m/s^2

Let's now calculate the ratio F/m. We have:

F = 29.4 N

m = 3.0 kg

Subsituting, we find

\frac{F}{m}=\frac{29.4}{3.0}=9.8 N/kg = 9.8 m/s^2

This is called acceleration of gravity, and it is the acceleration at which every object falls near the Earth's surface. It is indicated with the symbol g.

We can prove that this is the acceleration of the object: in fact, according to Newton's second law,

F=ma

where a is the acceleration of the object. Re-arranging,

a=\frac{F}{m}

which is exactly equal to the quantity we have calculated above.

5 0
3 years ago
Temperature change time base problem. Suppose V = 24 V, I = 0.1 A, for water: mw = 51 gm, cw = 4.18 J/gm ∘K-1, for resistor: mr
nirvana33 [79]

Answer:

t = 444.125 sec

Explanation:

Given data:

V = 24 volt

I  = 0.1 ampere

mass of water mw = 51 gm

cr = 4.18 J/gm degree K^-1

mass of resistor = 8 gm

cr = 3.7 J/gm degree K^-1

we know that power is given as

Power P = VI

But P =E/t

so equating both side we have

\frac{E}{t} = VI

solving for t

t = \frac{E}{VI}

t = \frac{m_w C_w \Delta T}{VI}

t = \frac{51 \times 4.18 \times (5 -0)}{24\times 0.1}

t = 444.125 sec

5 0
4 years ago
Read 2 more answers
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